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2 years ago

Write the equation in standard form using integers y=3x + 5

Mathematics

1 answer:

strojnjashka [21]2 years ago

4 0

Answer:

Put the equation into standard form

Ax+By=C

subtract 3x to the left

answer:

-3x+y=5

Ax+By=C

Step-by-step explanation:

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Determine which of the following formulas hold for all invertible n×n matrices A and B.

garik1379 [7]

Answer:

1 hold

2 hold

3 does not hold

4 hold

5 hold

6 hold

7 does not hold

8 does not hold

9 Does not hold

10 Hold

Step-by-step explanation:

The detailed step by step verification is as shown in the attachment

4 0

3 years ago

Use the distributive property to remove the parentheses. 9(x-3)

Svetllana [295]

The answer is 9x -27 because I distribute

4 0

2 years ago

How to calulate the area of a square with a perimeter of 60 inches

olasank [31]

Answer:

Step-by-step explanation:

The formula of an area of a square:

$A=s^2$

s - side

The formula of a perimeter of a square:

$P=4s$

s - side

We have a perimeter

$P=60\ in$

Substitute:

$4s=60$ divide both sides by 4

$\dfrac{4s}{4}=\dfrac{60}{4}\\\\s=15\ in$

Put it to the formula of an area of a square:

$A=15^2=225\ in^2$

6 0

2 years ago

Read 2 more answers

A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/mi

sergij07 [2.7K]

Answer:

(a) After time (t), the amount of salt in the tank is s(t) = 130 − 130e ∧−3t/200 / 3

(b) After 60 minutes, the salt in the tank is s(60) ≈ 25.7153

Step-by-step explanation:

To start with,

Let s(t) = amount of salt in kg of salt at time t.

Then we have:

dt/ds = (rate of salt into tank) − (rate of salt out of tank)

= (0.05 kg/L · 5 L/min) + (0.04 kg/L · 10 L/min) − ( s kg/ 1000 L x 15 L/min)

= 0.25 kg/min + 0.4 kg/min − 15s / 1000 kg/min

So we get the differential equation

dt/ds = 0.65 − 15s / 1000

= 65 / 100 − 15s / 1000

dt/ds = 130 - 3s / 200

We separate s and t to get

1 / 130 - 3s ds = 1 / 200 dt

Then we Integrate,

Thus we have

∫1 / 130 - 3s ds = ∫1 / 200 dt

- 1/3 · ln |130 − 3s| =1 / 200 t + C1

ln |130 − 3s| = - 3 /200 t + C2

|130 − 3s| = e− ³⁺²⁰⁰ ∧ t+C2

|130 − 3s| = C3e − ∧3t/200

130 − 3s = C4e ∧−3t/200

−3s = −130 + C4e ∧−3t/200

s = 130 - C4e ∧−3t/200 / 3

Since we begin with pure water, we have s(0) = 0. Substituting,

0 = 130 − C4e ∧−3·0/200 / 3

0 = 130 − C4

C4 = 130

So our function is

s(t) = 130 − 130e ∧−3t/200 / 3

After one hour (60 min), we have

s(60) = 130 − 130e ∧−3·60/200 / 3

s(60) ≈ 25.7153

Thus, after one hour there is about 25.72 kg of salt in the tank.

6 0

2 years ago

Andrew pays a monthly membership fee of $10 at his gym. Each time he uses the gym, he pays $5. Last month, Andrew spent a total

umka2103 [35]

Answer:

First step to solving the equation is to subtract 10 on both sides:

10 - 10 + 5x =65 -10

5x = 55

Then, we divide by 5 on both sides!

5x ÷ 5 = 55 ÷ 5

x = 11

Our result is 11, so the answer to the problem is 11

I hope I was helpful!

8 0

2 years ago